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A charged particle having charge q and mass m accelerated by a potential difference $V_{0}$ enters inside a parallel plate capacitor (parallel to length l of plates) at midpoint of its separation between plates at $t = 0$ . Potential difference across capacitor varies with time t as, $V = a d t$ , where a is constant and d is separation between plates.
Angle of deviation as it comes out of plates, is $\frac{π}{4}$ .
Value of $\frac{m a^{2} l^{4}}{2 q V_{0}^{3}}$ , will be

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Detailed Solution

$I = V_{x} t_{,} V_{x} = \sqrt{\frac{2 q V_{0}}{m}} V_{y} = \int \frac{q E}{m} d t = \frac{q}{m} \int a t d t$

$V_{y} = \frac{q}{m} a \frac{t^{2}}{2} V_{y} = \frac{q}{m} a \frac{l^{2}}{2 V_{x}^{2}} = \frac{q a l^{2}}{2 m} (\frac{m}{2 q V_{0}}) V_{y} = \frac{a l^{2}}{4 V_{0}} \tan θ = \frac{V_{y}}{V_{x}} = \frac{a l^{2}}{4} \sqrt{\frac{m}{2 q V_{0}^{3}}}$

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